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(C) Given,the latus rectum of the hyperbola subtends $90^{\circ}$ at its centre.Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.So,the ec

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$\frac{\sqrt{5}+1}{2}$

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(C) Given,the latus rectum of the hyperbola subtends $90^{\circ}$ at its centre.Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.So,the eccentricity is $e = \sqrt{1 + \frac{b^2}{a^2}}$,which implies $b^2 = a^2(e^2 - 1)$.The end points of the latus rectum are $L = (ae, \frac{b^2}{a})$ and $L' = (ae, -\frac{b^2}{a})$.The centre is $C = (0, 0)$.Since $\angle LCL' = 90^{\circ}$,the slopes of $CL$ and $CL'$ are $m_1 = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$ and $m_2 = \frac{-b^2/a}{ae} = -\frac{b^2}{a^2e}$.Since $CL \perp CL'$,we have $m_1 	imes m_2 = -1$.$\left(\frac{b^2}{a^2e}ight) 	imes \left(-\frac{b^2}{a^2e}ight) = -1$ $\Rightarrow \frac{b^4}{a^4e^2} = 1$ $\Rightarrow b^4 = a^4e^2$.Substituting $b^2 = a^2(e^2 - 1)$:$[a^2(e^2 - 1)]^2 = a^4e^2 \Rightarrow a^4(e^2 - 1)^2 = a^4e^2$.$(e^2 - 1)^2 = e^2 \Rightarrow e^2 - 1 = \pm e$.Case $1$: $e^2 - e - 1 = 0 \Rightarrow e = \frac{1 + \sqrt{5}}{2}$ (since $e > 1$).Case $2$: $e^2 + e - 1 = 0 \Rightarrow e = \frac{-1 + \sqrt{5}}{2}$ (rejected as $e > 1$).Thus,the eccentricity is $\frac{\sqrt{5} + 1}{2}$.

If the latus rectum subtends a right angle at the center of the hyperbola,then its eccentricity is

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